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Quick questions and answers

Old 11-19-05, 06:47 PM
  #1  
curlybro
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Quick questions and answers

I know I'm not a mod here, but this might alleviate some of the thread strain around these parts. In this thread you can ask quick questions that don't really warrant a thread. Something like "what do you guys think of *****". That would be a good question to ask in here. Also "do you guys use pedals on your bikes?" would also be appropriate. If this thread relieves the front page from two threads a day then it will be well worth keeping around.
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Old 11-19-05, 06:57 PM
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yes, what do you guys think of***** ?
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Old 11-19-05, 07:16 PM
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I really don't get why so many people are so anal about "unnecessary" or "redundant" threads.
Why can't people just ignore the threads that don't interest them and allow those who want to start their own thread to do so - regardless of the topic.
I don't understand the need to "relieve the front page from two threads a day."
So, what do you all think of that?
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Old 11-19-05, 07:23 PM
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Some people need to type less and ride more.

I tried to ride today, but the 4+ inches of snow put a damper on that.

I need some skis now...or a Puglsey!
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Old 11-19-05, 07:31 PM
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So what bag goes with a pugsley?
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Old 11-19-05, 09:56 PM
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paper or plastic?
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Old 11-20-05, 02:55 AM
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yeah, just ignore the threads you don't like...

i'm not sure how to put this philosophically...people waste their time writing replies because they CAN... just like i'm doing now.
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Old 11-20-05, 03:27 AM
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Threads that you have no knowledge to contribute to or have anothing to do with... DO NOT REPLY.

Easy!

ps. yes... kinda like what I just did... don't do that!
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Old 11-20-05, 03:30 AM
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Check out my new "I Already Knew That" thread. It seeks to alieviate some of the stress of the "Quick Question" thread by inviting people to shed light on the fact that they already know the answer to whatever question you intend to ask (as oppoesed to actually answering 'em). Perhaps the mods will make it a "sticky."
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Old 11-20-05, 04:06 PM
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I've always had some quick questions in mind, now i have chance to ask them:

• If a stripper gets breast implants can she write it off on her taxes as a business expense?
• How do "do not walk on grass" signs get there?
• If there was a crumb on the table and you cut it in half, would you have two crumbs or two halves of a crumb?
• Do illiterate people get the full effect of Alphabet soup?
• In that song, she'll be coming around the mountain, who is she?
and lastly
• Why do all superheroes wear spandex?
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Old 11-20-05, 04:26 PM
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Originally Posted by ostro
I've always had some quick questions in mind, now i have chance to ask them:

• If a stripper gets breast implants can she write it off on her taxes as a business expense?
• How do "do not walk on grass" signs get there?
• If there was a crumb on the table and you cut it in half, would you have two crumbs or two halves of a crumb?
• Do illiterate people get the full effect of Alphabet soup?
• In that song, she'll be coming around the mountain, who is she?
and lastly
• Why do all superheroes wear spandex?
  • yes
  • the rule doesn't go into effect until the feet of the installer are off the grass
  • two crumbs
  • no
  • sally the mule (don't ask)
  • because it's easy to get stains out of spandex

next.
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Old 11-20-05, 04:48 PM
  #12  
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ok...

• If you die and you have a broken leg do they take the cast off?
• Since there is a rule that states "i" before "e" except after "c", isnt "science" spelled wrong?
• Why isn't the word 'gullible' in the dictionary?
•*Why did Superman wear his briefs on the outside of his tights?
• How many licks does it take to get to the center of a tootsie pop?
• Why do British people never sound British when they sing?
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Old 11-20-05, 04:55 PM
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Originally Posted by ostro
ok...

• If you die and you have a broken leg do they take the cast off?
• Since there is a rule that states "i" before "e" except after "c", isnt "science" spelled wrong?
• Why isn't the word 'gullible' in the dictionary?
•*Why did Superman wear his briefs on the outside of his tights?
• How many licks does it take to get to the center of a tootsie pop?
• Why do British people never sound British when they sing?
  • yes
  • it's spelled correctly, but english doesn't follow rules very well
  • nice try
  • 'cause he's an alien
  • no one knows, because you always crunch them up before you get there
  • they do sound british, it just so happens that recieved pronunciation sounds north american when sung.
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Old 11-20-05, 04:57 PM
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How many cyclists can legaly ride abrest?
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Old 11-20-05, 04:58 PM
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Originally Posted by JoshFrank
How many cyclists can legaly ride abrest?
that's 'abreast' smart guy
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Old 11-20-05, 05:02 PM
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oops, to quick with my bad pun...
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Old 11-20-05, 05:04 PM
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damn, the quick question thread works.

What is the geometrical meaning of the central extension of the algebra of diffeomorphisms of a circle?
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Old 11-20-05, 05:27 PM
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of the whole group, or for a given circle?
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Old 11-20-05, 05:54 PM
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Originally Posted by dolface
of the whole group, or for a given circle?
uhhh...yeah...a given circle. Yeahup thats it, for a circle.
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Old 11-20-05, 06:16 PM
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ok, i haven't looked at the problem in a long time, so all i've got is a general explanation (you can plug in the numbers for whatever circle you have).

(if you don't care about the math [which sorta makes it pointless] this is a very basic description: there is a lie algebra called the virasoro algebra; its almost just the lie algebra of the group of diffeomorphisms of the circle, but it's actually just one dimension bigger, being a "central extension" thereof; projective representations of the lie algebra of the group of diffeomorphisms of the circle correspond to honest representations of the virasoro algebra.)

i'm not gonna re-type my notes, but this post (from google groups) covers it pretty well:

"In 2D the conformal group is infinite-dimensional - it
consists of transformations which are analytic in z = x+iy and z* =
x-iy. If we concentrate on the former, we see that an infinitesimal
conformal transformation is generated by

L_m = z^{m+1} d/dz,

and they obey the algebra

[L_m, L_n] = (n-m) L_{m+n}.

However, it is easy to see that this is the algebra of vector fields
(= infinitesimal diffeomorphisms) on the circle, vect(1). If the
circle coordinate is x, the generators are

L_m = -i exp(imx) d/dx.

vect(1) has a central extension, known as the Virasoro algebra:

[L_m, L_n] = (n-m) L_{m+n} - c/12 (m^3 - m) delta_{m+n,0},

where c is a c-number known as the central charge or conformal
anomaly. This means that the Virasoro algebra is still a Lie
algebra - anti-symmetry and the Jacobi identities still hold. The
term linear in m is unimportant, because it can be removed by a
redefinition of L_0.

The m^3 term is a genuine quantum effect, which simply is there when
you quantize a string. When string theorists criticize Thiemann's
LQG string, they are basically complaining that he does not get
this term, which simply must be there.

There is some confusion about anomaly freedom here, because at the
end people want to eliminate the conformal symmetry. The nice way to
do this is to introduce a ghost pair b_m, c_n, satisfying fermionic
brackets

{ b_m, c_n } = delta_{m+n,0}.

One can now write down the BRST operator, which looks something
like (double dots denote normal ordering)

Q = sum_m :L_{-m} c_m: + 1/2 sum_{m,n} (m-n) :b_{-m-n} c_m c_n:

If the BRST operator is nilpotent, Q^2 = 0, we can identify
physical states with BRST co****logy. A state is physical if it
is BRST closed, Q|phys> = 0, and two states are equivalent if
they differ by a BRST exact term,

|phys> ~ |phys'> if Q( |phys> - |phys'> ) = 0.

It turns out that the ghost has central charge c = -26, so the BRST
operator is nilpotent if the L_m's have c = 26; this is where the 26
dimensions of they bosonic string comes from. However, the important
thing from my viewpoint is not that the end result is anomaly free,
but that an anomaly exists, even if only in intermediate
calculations. Thiemann does not have an anomaly even intermediately,
and from this string theorists (and myself) conclude that LQG is
wrong.

Let us now return to the math. A lowest-weight representation (LWR)
of the Virasoro algebra is characterized by a lowest-weight state
|h,c> satisfying

L_0 |h,c> = h |h,c>,

L_{-m} |h,c> = 0, for all -m < 0.

It is known that the only unitary LWR of vect(1) is the trivial one.
However, the Virasoro algebra has many unitary LWRs: the discrete
series with 0 <= c < 1, where

c = 1 - 6/m(m+1), m positive integer and

h = h_{p,q}(c) = (pm^2 - q(m+1)^2) / 4m(m+1)

(or something similar, I'm quoting from memory), and also all c > 1,
h > 0. Anyway, the important thing is that the only acceptable value
of (h,c) with c = 0 is h = 0 - this is the trivial representation.
That Thiemann obtains a non-trivial unitary representation of the
1D diffeomorphism group with c = 0 is thus very strange. It is
hard to see how that could be compatible with quantum theory.

The Virasoro algebra can be generalized to several dimensions -
an extension of the diffeomorphism algebra on the N-dimensional
torus, say. The generators are

L_k(m) = i exp(im.x) d/dx^k,

where m = (m_i) and m.x = m_i x^i and I use the summation
convention. The algebra depends on two parameters (abelian charges)
c_1 and c_2,

[L_i(m), L_j(n)] = n_i L_j(m+n) - m_j L_i(m+n)

+ (c_1 m_j n_i + c_2 m_i n_j) m_k S^k(m+n),

[L_i(m), S^j(n)] = n_i S^j(m+n) + delta^j_i m_k S^k(m+n),

[S^i(m), S^j(n)] = 0,

m_k S^k(m) = 0.

This is the Virasoro algebra in 1D because the last condition then
becomes m S(m) = 0, which only has the solution S(m) ~ delta(m).
The Virasoro extension is not central (does not commute with
everything) except in 1D. It is straightforward but somewhat tedious
to check that these relations indeed do define a Lie algebra.

Just as the Virasoro algebra is anomalous when c != 0, its higher-
dimensional sibling is anomalous unless both c_1 = c_2 = 0. And just
as for the Virasoro algebra, there is no non-trivial, unitary LWRs
unless the algebra is anomalous. The correct definition of
lowest-weight is more subtle in several dimensions. Let me just say
that the right definition does not introduce any anisotropy."

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Old 11-20-05, 06:58 PM
  #21  
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I <heart> dolface.
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Old 11-20-05, 07:00 PM
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*point - dolface*
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Old 11-20-05, 07:09 PM
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dol, that made me nauseous.
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Old 11-20-05, 07:17 PM
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Originally Posted by steaktaco
dol, that made me nauseous.
me too, and i think i hurt my brain going over that stuff again.
i need another beer
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Old 11-20-05, 07:23 PM
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dolface, if you didnt copy/paste that, you need to get out on your bike
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